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I have two particle detectors (random neutron source), for each I know the arrival waiting time distribution $PDF_1(t)$ and $PDF_2(t)$. They are non-exponential, since the detectors have dead-time, and their dead-times are different. Nevertheless, $PDF_1(t)$ and $PDF_2(t)$ are known.

I would like to calculate the waiting time distribution when I combine the signal from both detectors, given that the signal merging does not introduce any additional dead-time.

I know it is simple for waiting time for Poisson processes: if $PDF_1(t)=R_1 * \exp(-R_1 * t)$ and $PDF_2(t)=R_2 * \exp(-R_2 * t)$, the waiting time for the combined signal would be $(R_1+R_2)*\exp(-(R_1+R_2)*t)$

But I'd like to calculate it for arbitrary waiting time PDFs numerically. If someone can give a hint how I can do it: a general formula or list of steps to follow..?

Vova N
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    Regardless of what the two PDFs are, as long as the waiting times they model are independent the PDF of the sum of those waiting times is the convolution. – Kurt G. Mar 16 '23 at 19:22
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    Thank you for your answer. I think that's not correct, e.g. if both waiting time PDFs are exponents with the same decay constant, the convolution would give Gamma distribution with shape parameter k=2. While the combined signal waiting time PDF should remain exponential with doubled decay constant.. Thanks, anyway! – Vova N Mar 16 '23 at 19:40
  • I agree now that my comment did not address your question but I also realise that the reason for this confusion is that you seem to mix up wating times with the counting process. Poisson waiting times are exponential, jump times are Gamma and the Poisson process itself is Poisson. The sum of two independent Poissons is Poisson with intensity $\lambda_1+\lambda_2,.$ Please rethink this and edit the question. – Kurt G. Mar 16 '23 at 19:46

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